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When one object collides with another object of the same mass in a 2D plane, we know that we can derive that the angles that the objects leave the collision at add up to 90 degrees in a perfectly elastic collision.

However, when one of these angles is 0, the angle magically becomes 180 degrees or 0 degrees instead of 90 degrees.

This is intuitive; if a ball hits another ball at no angle, you would expect it to bounce off at 180 degrees, or in the opposite direction. However, why is this the case that only when one angle is 0, the formula does not hold? What happens as that angle approaches 0?

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Hint: One can include the mentioned special cases by instead stating that the dot product of the two final velocities is zero. This follows from momentum and energy conservation. (Here we have assumed an elastic non-relativistic collision of two equal masses with one mass initially at rest.) If the two final velocities are both different from zero, then the angle must be $90^{\circ}$.

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  • $\begingroup$ It's once in a blue moon that you answer queries on Newtonian Mechanics! At least since my joining I've got the opportunity to see an answer from you on Newtonian Mechanics only now! Well done, sir;+1:) $\endgroup$ – user36790 Oct 11 '15 at 4:28
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If the velocity of one particle is zero after the collision then the direction in which it is travelling is ambiguous. The magnitude of velocity is zero, but it can be in any direction.

In such situations you need to look at the limiting case. As you reduce the magnitude of velocity of particle #1 towards zero, the direction of particle #2 becomes closer to the the initial direction, while the direction of particle #1 remains perpendicular to #2. Velocity vector #1 vanishes but as it does so its limiting direction remains perpendicular to the velocity vector #2.

So the case in which the incoming particle comes to rest while the stationary particle moves off in the initial direction is not a separate case, it is in fact covered by the same formula as a limiting case.

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